Maximum Product

June 11, 2019

There are four ways the maximum product can be formed:

1) take the maximum from each of the three arrays
2) take the maximum from the first array and the minimum from the second and third arrays
3) take the maximum from the second array and the minimum from the first and third arrays
4) take the maximum from the third array and the minimum from the first and second arrays:

It is not correct, as one commenter suggested, to sort all three arrays, sort the six-element array built as the first and last item from each sorted array, and then take either the three maximum items or the maximum plus two minimum items from the six-element array, as that does not require all three elements to come from three different arrays.

Note that to avoid unnecessary iterations, I limited the total number of iteration by including the iwp threshold (iterations without progress). This is heuristically calculated using the sizes of the original arrays as an input. Tried it with larger arrays and it works, though I cannot vouch for its effectiveness for all sizes. Cheers!

@Zack: I don’t understand how the ‘iwp’ cutoff is correct. If there is a very large item at the end of array C, won’t it be missed? Also, your method seems to be Theta(n^3) instead of the other Theta(n) solutions.

@Paul: that’s a good question. That’s why I sort the arrays first, in terms of absolute values. Chances are that the first iterations of the loops are going to yield the optimum result, so we never have to go through all possible combinations (something wasteful for large arrays). The latter are more than 8 for the original arrays, btw; namely Total_Combinations = 4 x 4 x 4 = 64. Cheers